An electric cable of radius r_l and thermal conductivity k_e is enclosed by an insulating sleeve whose outer surface is of radius r₂ and experiences convection heat transfer and radiation exchange with the adjoining air and large surroundings, respectively. When electric current passes through the cable, thermal energy is generated within the cable at a volumetric rate q.

(a) Write the steady-state forms of the heat diffusion equation for the insulation and the cable. Verify that these equations are satisfied by the following temperature distributions: Sketch the temperature distribution, T(r), in the cable and the sleeve, labeling key features.

(b) Applying Fourier's law, show that the rate of conduction heat transfer per unit length through the sleeve may be ex pressed as qr = 2πK_s (T_s,1 – T_s,2) /In r₂/r₁) Applying an energy balance to a control surface placed around the cable obtain an alternative expression for q; expressing your result in terms of q and r).

(c) Applying an energy balance to a control surface placed around the outer surface of the sleeve obtain an expression from which T_s.2 may be determined as a function of q, r_l, h, T_∞, ε, and T_sur.

(d) Consider conditions for which 250 A are passing through a cable having an electric resistance per unit length of R/ = 0.005 film, a radius of r _l = 15 mm, and a thermal conductivity of kc = 200 W/m' K.

For ks = 0.15 W/m. K, r2 = 15.5 mm, h = 25 W/m² ∙ K, ϵ = 0.9, T_∞ = 25°C, and T_sur = 35°C, evaluate the surface temperatures, T_s.1 and T_s.2, as well as the temperature T_o at the centerline of the cable.

(e) With all other conditions remaining the same, compute and plot T_o, T_s.1, and T_s.2 as a function of r₂ for 15.5 ≤ r₂ ≤ 5 20 mm.

Transcribed Image Text:

Electrical cable Insulation Ambient air 11 In(rir) - T): In[nfr) Insulation: Tr) = T, + (T Cable: T(r) = T+